Problem: You have found the following ages (in years) of all 4 seals at your local zoo: $ 5,\enspace 13,\enspace 13,\enspace 4$ What is the average age of the seals at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 4 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{5 + 13 + 13 + 4}{{4}} = {8.8\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $5$ years $-3.8$ years $14.44$ years $^2$ $13$ years $4.2$ years $17.64$ years $^2$ $13$ years $4.2$ years $17.64$ years $^2$ $4$ years $-4.8$ years $23.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{14.44} + {17.64} + {17.64} + {23.04}} {{4}} $ $ {\sigma^2} = \dfrac{{72.76}}{{4}} = {18.19\text{ years}^2} $ The average seal at the zoo is 8.8 years old. The population variance is 18.19 years $^2$.